Answer:
The concentration of methyl isonitrile will become 15% of the initial value after 10.31 hrs.
Explanation:
As the data the rate constant is not given in this description, However from observing the complete question the rate constant is given as a rate constant of 5.11x10-5s-1 at 472k .
Now the ratio of two concentrations is given as

Here C/C_0 is the ratio of concentration which is given as 15% or 0.15.
k is the rate constant which is given as 
So time t is given as

So the concentration will become 15% of the initial value after 10.31 hrs.
Answer: The approximate molecular mass of the polypeptide is 856 g/mol
Explanation:
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
Or,
where,
= osmotic pressure of the solution = 4.19 torr
i = Van't hoff factor = 1 (for non-electrolytes)
Mass of solute (polypeptide) = 0.327 g
Volume of solution = 1.70 L
R = Gas constant =
T = temperature of the solution =
Putting values in above equation, we get:
Hence, the molar mass of the polypeptide is 856 g/mol
4) is correct
This is because water is polar and it will mix with a polar solvent. A good rule for remembering the behavior of non-polar and polar compounds when it comes to being miscible is that "like dissolves like."
Pb(NO₃)₂ ⇒limiting reactant
moles PbI₂ = 1.36 x 10⁻³
% yield = 87.72%
<h3>Further explanation</h3>
Given
Reaction(unbalanced)
Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)
Required
- moles of PbI₂
- Limiting reactant
- % yield
Solution
Balanced equation :
Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)
mol Pb(NO₃)₂ :
= 0.45 : 331 g/mol
= 1.36 x 10⁻³
mol NaI :
= 250 ml x 0.25 M
= 0.0625
Limiting reactant (mol : coefficient)
Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³
NaI : 0.0625 : 2 = 0.03125
Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)
moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)
Mass of PbI₂ :
= mol x MW
= 1.36 x 10⁻³ x 461,01 g/mol
= 0.627 g
% yield = 0.55/0.627 x 100% = 87.72%